Question

Patients with recurring ventricular arrhythmia, irregular beats of the heart's ventricles, are often fitted with implantable cardioverter defibrillators (ICD) that monitor the beating of the patient's heart. Much like an external defibrillator, an ICD delivers up to 40.0 J of energy to the heart when irregular heartbeat is detected. A standard ICD contains a 3.2 V lithium ion battery and has an equivalent capacitance of 140 µF.

1. What is the charge (in C) stored on the fully charged capacitor in an ICD?
2. What should the internal resistance (in Ω) of the charging circuit in the ICD be if the capacitor is to attain 93.2% of its maximum charge within 1.80 s?
3. The total effective resistance of the circuit when the ICD discharges through the heart is 500 Ω, in series with the capacitor in the ICD. During what time interval (in ms) would the fully charged ICD deliver 63% of its stored energy to the heart?

Answer 1

a)

Charge

Q=CV =(140*10^-6)(3.2) =4.48*10^-4 C

b)

In a RC circuit charge as a function of time is given by

Q=Q_max(1-e^-t/T)

0.932Q_max = Q_max ​​​​​​​(1-e^-1.8/T)

Ln(1-0.932) = -1.8/T

T=0.66958 s

Since T=RC

R=0.66958/(140*10^-6)

R=4782.72 ohms or 4.783 Kohms

c)

Time Constant

T=RC=500*140*10^-6 = 0.07 s

For discharging

Q=Q_maxe^-t/T

0.63Q_max = Q_maxe^-t/0.07

Ln(0.63) =-t/0.07

t=0.03234 s